Double Asymptotic for Random Walks on Hypercubes

Abstract

We consider the sum of the coordinates of a simple random walk on the K-dimensional hypercube and prove a double asymptotic of this process, as both the time parameter n and the space parameter K tend to infinity. Depending on the asymptotic ratio of the two parameters, the rescaled processes converge toward either a “stationary Brownian motion,” an Ornstein–Uhlenbeck process or a Gaussian white noise.

Appendix 1: Proof of Lemma 1

Recall Lemma 1 which comes from the “Fast regime” section:

Lemma 1

Let $\left( \xi _i\right) _{i\in {\mathbb {N}}}$ be i.i.d. Rademacher random variables and $\left( \mathbf{t }_K\right) _{K\ge 1}$ a sequence of random vectors in $\left( {\mathbb {R}} _+\right) ^k$ independent from $\left( \xi _i\right) _{i\in {\mathbb {N}}}$. Define for all $K\ge 1$ and $\mathbf{s }=(s_1,\ldots ,s_k)\in \left( {\mathbb {R}} _+\right) ^k$:

$$\begin{aligned} T_K(\mathbf{s })=\left( \frac{1}{\sqrt{K}}\sum _{i=1}^{\lfloor Ks_j\rfloor }\xi _i\right) _{j\in [k]}. \end{aligned}$$

If there exists a deterministic $\mathbf{t }=(t_1,\ldots ,t_k)\in \left( {\mathbb {R}} _+\right) ^k$ such that $\mathbf{t }_K\overset{{\mathbb {P}}}{\underset{K\rightarrow \infty }{\longrightarrow }}\mathbf{t }$, then:

$$\begin{aligned} T_K(\mathbf{t }_K)\overset{\mathcal {D}}{\underset{K\rightarrow \infty }{\longrightarrow }}\left( W_{t_j}\right) _{j\in [k]} \end{aligned}$$

where $\left( W_s\right) _{s\in {\mathbb {R}} _+}$ denotes a standard Brownian motion starting from 0.

For all $\mathbf{x }=(x_1,\ldots ,x_k)\in {\mathbb {R}} ^k$ we will consider the following norm:

$$\begin{aligned} \vert \vert \mathbf{x }\vert \vert = \sup _{i\in [k]} \vert x_i\vert . \end{aligned}$$

For the sake of simplicity we will set ${\mathbb {R}} _+^k\overset{\mathrm{def}}{=}\left( {\mathbb {R}} _+\right) ^k$, and for every ${\mathbb {R}} ^k$-valued process $\left( X_t\right) _{t\ge 0}$ and $\mathbf{t }=(t_1,\ldots ,t_k)\in {\mathbb {R}} _+^k$, we will note $X_\mathbf{t }\overset{\mathrm{def}}{=}\left( X_{t_1},\ldots ,X_{t_k}\right)$.

In order to prove Lemma 1 we will use several times the following result:

Lemma 4

Let $\left( W_t\right) _{t\in {\mathbb {R}} _+}$ be a standard Brownian motion starting from 0. Then for all $\varepsilon >0$ and $d>0$ there exists $\delta >0$ such that:

$$\begin{aligned} \vert \vert \mathbf{x }-\mathbf{y }\vert \vert \le \delta \ \Rightarrow \ {\mathbb {P}} (\vert \vert W_\mathbf{x }-W_\mathbf{y }\vert \vert > d)< \varepsilon . \end{aligned}$$

Proof

$$\begin{aligned} {\mathbb {P}} (\vert \vert W_\mathbf{x }-W_\mathbf{y }\vert \vert> d)&= {\mathbb {P}} \left( \sup _{1\le i\le k} \vert W_{x_i}-W_{y_i} \vert> d\right) \\&\le \sum _{i=1}^k {\mathbb {P}} (\vert W_{x_i}-W_{y_i} \vert> d) \\&= \sum _{i=1}^k {\mathbb {P}} (\vert N_i \vert > d) \end{aligned}$$

where $N_i$ is a $\mathcal {N}(0,\vert x_i-y_i\vert )$ random variable. One can easily find a suitable $\delta >0$ such that:

$$\begin{aligned} \vert x_i-y_i\vert \le \delta \ \Rightarrow \ {\mathbb {P}} (\vert N_i\vert > d)< \frac{\varepsilon }{k} \end{aligned}$$

and get the result using the fact that $\vert \vert \mathbf{x }-\mathbf{y }\vert \vert \le \delta \Rightarrow \vert x_i-y_i\vert \le \delta$ for all $i\in [k]$. $\square$

Let us get back to the main proof. Assuming the premises of Lemma 1 we want to prove that for every function $f:{\mathbb {R}} _+^k\rightarrow {\mathbb {R}}$ bounded and uniformly continuous we have:

$$\begin{aligned} {\mathbb {E}} [f(T_K(\mathbf{t }_K))]\underset{K\rightarrow \infty }{\longrightarrow }{\mathbb {E}} [f(W_\mathbf{t })]. \end{aligned}$$

Let $\varepsilon >0$. For all $\delta > 0$ we have:

Since $\mathbf{t }_K\overset{{\mathbb {P}}}{\underset{K\rightarrow \infty }{\longrightarrow }}\mathbf{t }$ and then $\forall \ \delta >0$ there exists $N_1(\varepsilon ,\delta )\in {\mathbb {N}}$ such that for every $K\ge N_1(\varepsilon ,\delta )$ we have:

$$\begin{aligned} {\mathbb {P}} (\vert \vert \mathbf{t }_K-\mathbf{t }\vert \vert >\delta ) \le \frac{\varepsilon }{8\vert \vert f\vert \vert _\infty }. \end{aligned}$$

(9)

Now we split the other term in two parts (which will be dominated separately):

Let us define the functions $\varphi _K$ and $\varphi$ by:

$$\begin{aligned} \varphi _K(\mathbf{s })={\mathbb {E}} [f(T_K(\mathbf{s }))] \quad \text {and}\quad \varphi (\mathbf{s })={\mathbb {E}} [f(W_\mathbf{s })] \ \text {for } \mathbf{s }\in {\mathbb {R}} _+^k. \end{aligned}$$

Proposition 5

The functions $\left( \varphi _K\right) _{K\ge 1}$ converge uniformly on every compact of ${\mathbb {R}} _+^k$ to the function K.

Proof

We already know via the Donsker’s theorem that the sequence $\left( \varphi _K\right) _{K\ge 1}$ converges pointwise to $\varphi$ (see, for instance, [6]).

Let $S\subset {\mathbb {R}} _+^k$ compact; for all $\delta >0$, there exists a finite subset $\mathcal {M}\subset S$ such that:

$$\begin{aligned} \forall \ \mathbf{x }\in S \ \exists \ \mathbf{y }\in \mathcal {M}: \vert \vert \mathbf{x }-\mathbf{y }\vert \vert \le \delta . \end{aligned}$$

Let $\varepsilon >0$ and $\mathbf{t }\in S$, and choose $\mathbf{s }\in \mathcal {M}$ such that $\vert \vert \mathbf{t }-\mathbf{s }\vert \vert \le \delta$. We then get:

$$\begin{aligned} \vert \varphi _K(\mathbf{t })-\varphi (\mathbf{t })\vert \le \vert \varphi _K(\mathbf{t })-\varphi _K(\mathbf{s })\vert + \vert \varphi _K(\mathbf{s })-\varphi (\mathbf{s })\vert + \vert \varphi (\mathbf{s })-\varphi (\mathbf{t })\vert . \end{aligned}$$

(10)

Let us dominate the first term. The function f being uniformly continuous, there exists $d >0$ such that $\vert \vert \mathbf{x }-\mathbf{y }\vert \vert \le d \Rightarrow \vert f(\mathbf{x })-f(\mathbf{y })\vert < \varepsilon /6$. Then we get:

But we can see in the proof of Theorem 4.20 (p. 70) from [6] that $\forall \ c >0$ et $\forall \ D>0$:

$$\begin{aligned} \lim _{\delta \rightarrow 0} \sup _{n\ge 1} {\mathbb {P}} \left( \max _{\begin{array}{c} \vert x-y\vert \le \delta \\ 0\le x,y \le D \end{array}} \left| T_K(x)-T_K(y)\right| >c\right) =0. \end{aligned}$$

So there exists some $\delta _1 (\varepsilon ,d)>0$ such that $\forall \ \delta \le \delta _1 (\varepsilon ,d)$:

$$\begin{aligned} \sup _{n\ge 1} {\mathbb {P}} \left( \max _{\begin{array}{c} \vert x-y\vert \le \delta \\ 0\le x,y \le D \end{array}} \left| T_K(x)-T_K(y)\right| >d\right) \le \frac{\varepsilon }{12\vert \vert f\vert \vert _\infty k} \end{aligned}$$

and then, setting $D(S)\overset{\mathrm{def}}{=}\underset{\mathbf{z }\in S}{\sup }\vert \vert \mathbf{z }\vert \vert$, if we choose $\delta$ smaller than $\delta _1 (\varepsilon ,d)$ we have $\forall \ K\ge 1$:

$$\begin{aligned} {\mathbb {P}} (\vert \vert T_K(\mathbf{t })-T_K(\mathbf{s })\vert \vert> d)&\le k \times {\mathbb {P}} \left( \max _{\begin{array}{c} \vert x-y\vert \le \delta \\ 0\le x,y \le D(S) \end{array}} \left| T_K(x)-T_K(y)\right| >d\right) \\&\le \frac{\varepsilon k}{12\vert \vert f\vert \vert _\infty k}=\frac{\varepsilon }{12\vert \vert f\vert \vert _\infty }. \end{aligned}$$

We finally get:

$$\begin{aligned} \vert \varphi _K(\mathbf{t })-\varphi _K(\mathbf{s })\vert \le {}&\frac{\epsilon }{6}{\mathbb {P}} (\vert \vert T_K(\mathbf{t })-T_K(\mathbf{s })\vert \vert \le d) \\&+2\vert \vert f \vert \vert _\infty {\mathbb {P}} (\vert \vert T_K(\mathbf{t })-T_K(\mathbf{s })\vert \vert > d) \\ \le {}&\frac{\epsilon }{6} + 2\vert \vert f \vert \vert _\infty \frac{\varepsilon }{12\vert \vert f\vert \vert _\infty } =\frac{\varepsilon }{3}. \end{aligned}$$

Let us deal with the second term in (10), namely $\vert \varphi _K(\mathbf{s })-\varphi (\mathbf{s })\vert$. Since $\varphi _K$ converges pointwise to $\varphi$, for every $\mathbf{x }\in {\mathbb {R}} _+^k$ there exists $N_\mathbf{x }(\varepsilon )$ such that:

$$\begin{aligned} K\ge N_\mathbf{x }(\varepsilon ) \Longrightarrow \vert \varphi _K(\mathbf{x })-\varphi (\mathbf{x })\vert \le \frac{\varepsilon }{3}. \end{aligned}$$

We just have to take K greater than $N(\varepsilon ,\delta )\overset{\mathrm{def}}{=}\underset{\mathbf{x }\in \mathcal {M}}{\max }\ N_\mathbf{x }(\varepsilon )$ to get:

$$\begin{aligned} \vert \varphi _K(\mathbf{s })-\varphi (\mathbf{s })\vert \le \frac{\varepsilon }{3}. \end{aligned}$$

For the third term in (10) we use again the uniform continuity of f to get the following:

Using Lemma  we find that $\exists \ \delta _2(\varepsilon ,d)$ such that:

$$\begin{aligned} \delta \le \delta _2(\varepsilon ,d) \Longrightarrow {\mathbb {P}} (\vert \vert W_\mathbf{s }-W_\mathbf{t }\vert \vert > d) <\frac{\varepsilon }{12\vert \vert f\vert \vert _\infty }. \end{aligned}$$

We then have:

$$\begin{aligned} \vert \varphi (\mathbf{s })-\varphi (\mathbf{t })\vert\le & {} \frac{\varepsilon }{6}{\mathbb {P}} (\vert \vert W_\mathbf{s }-W_\mathbf{t }\vert \vert \le d) + 2\vert \vert f\vert \vert _\infty {\mathbb {P}} (\vert \vert W_\mathbf{s }-W_\mathbf{t }\vert \vert > d)\\\le & {} \frac{\varepsilon }{6} + 2\vert \vert f\vert \vert _\infty \frac{\varepsilon }{12\vert \vert f \vert \vert _\infty } =\frac{\varepsilon }{3}. \end{aligned}$$

Grouping all these results in (10) we get that if $\delta \le \min (\delta _1(\varepsilon ,d),\delta (\varepsilon ,d))$ then $\forall \ K\ge N(\varepsilon ,\delta )$ we have:

$$\begin{aligned} \sup _{\mathbf{t }\in S}\vert \varphi _K(\mathbf{t })-\varphi (\mathbf{t })\vert \le \frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3}=\varepsilon . \end{aligned}$$

$\square$

Using the previous proposition we get that for every S compact subset of ${\mathbb {R}} _+^k$ there exists $N(\varepsilon ,S)$ such that $\forall \ \mathbf{s }\in S$ and $\forall \ K\ge N(\varepsilon ,S)$:

$$\begin{aligned} \vert \varphi _K(\mathbf{s })-\varphi (\mathbf{s })\vert \le \frac{\varepsilon }{4}. \end{aligned}$$

Using the independence of the $(\mathbf{t }_K)_K$ and the $(\xi _i)_i$ we can write that:

$$\begin{aligned} {\mathbb {E}} [f(T_K(\mathbf{t }_K))]&= {\mathbb {E}} [{\mathbb {E}} [f(T_K(\mathbf{t }_K))\vert \mathbf{t }_K]] = {\mathbb {E}} [\varphi _K(\mathbf{t }_K)]. \end{aligned}$$

Then, since the ball of radius $\delta$ centered on $\mathbf{t }$ is compact, there exists an integer $N_2(\varepsilon ,\delta )\overset{\mathrm{def}}{=}N(\varepsilon ,B(\mathbf{t },\delta ))$ such that for all $K\ge N_2(\varepsilon ,\delta )$ we have:

(11)

Now we have to dominate . Using the uniform continuity of f we know there exists some $d>0$ such that:

$$\begin{aligned} \vert \vert \mathbf{x }-\mathbf{y }\vert \vert \le d \Rightarrow \vert f(\mathbf{x })-f(\mathbf{y })\vert \le \frac{\varepsilon }{4}. \end{aligned}$$

We then have:

(12)

Using Lemma , for all $d>0$ there exists $\delta (\varepsilon , d) >0$ small enough such that the following holds:

$$\begin{aligned} {\mathbb {P}} (\vert \vert \mathbf{t }_K-\mathbf{t }\vert \vert \le \delta ,\vert \vert W_{\mathbf{t }_K}-W_\mathbf{t }\vert \vert > d)\le \frac{\varepsilon }{8\vert \vert f \vert \vert _\infty }. \end{aligned}$$

(13)

In a nutshell if we sum up all the previous step, we find that for every bounded continuous function f and $\forall \ \varepsilon >0$, we can choose $d\le d(\varepsilon )$ such that (12) holds, $\delta \le \delta (\varepsilon ,d)$ to have (13), and then, $K\ge \max (N_1(\varepsilon ,\delta ),N_2(\varepsilon ,\delta )$ to get (9) and (11), which finally yields:

and thus $T_K(\mathbf{t }_{\mathbf{K }}) \overset{\mathcal {D}}{\underset{K\rightarrow \infty }{\longrightarrow }}W_\mathbf{t }$.